A question about special linear group
$begingroup$
Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$
linear-algebra matrices orthogonal-matrices
asked 1 hour ago
TotoroTotoro
1417
$endgroup$
add a comment |
$begingroup$
Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$
linear-algebra matrices orthogonal-matrices
asked 1 hour ago
TotoroTotoro
1417
$endgroup$
$begingroup$
Yes, these are matrices with finite order. It's an exercise, not really research level.
$endgroup$
– YCor
1 hour ago
add a comment |
$begingroup$
Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$
linear-algebra matrices orthogonal-matrices
asked 1 hour ago
TotoroTotoro
1417
$endgroup$
Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$
linear-algebra matrices orthogonal-matrices
linear-algebra matrices orthogonal-matrices
asked 1 hour ago
TotoroTotoro
1417
asked 1 hour ago
TotoroTotoro
1417
asked 1 hour ago
TotoroTotoro
1417
asked 1 hour ago
TotoroTotoro
1417
asked 1 hour ago
TotoroTotoro
1417
1417
$begingroup$
Yes, these are matrices with finite order. It's an exercise, not really research level.
$endgroup$
– YCor
1 hour ago
add a comment |
$begingroup$
Yes, these are matrices with finite order. It's an exercise, not really research level.
$endgroup$
– YCor
1 hour ago
$begingroup$
Yes, these are matrices with finite order. It's an exercise, not really research level.
$endgroup$
– YCor
1 hour ago
$begingroup$
Yes, these are matrices with finite order. It's an exercise, not really research level.
$endgroup$
– YCor
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
These are exactly the finite-order elements:
If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.
Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.
answered 1 hour ago
Andy PutmanAndy Putman
31.6k7134214
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
These are exactly the finite-order elements:
If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.
Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.
answered 1 hour ago
Andy PutmanAndy Putman
31.6k7134214
$endgroup$
add a comment |
$begingroup$
These are exactly the finite-order elements:
If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.
Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.
answered 1 hour ago
Andy PutmanAndy Putman
31.6k7134214
$endgroup$
add a comment |
$begingroup$
These are exactly the finite-order elements:
If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.
Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.
answered 1 hour ago
Andy PutmanAndy Putman
31.6k7134214
$endgroup$
These are exactly the finite-order elements:
If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.
Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.
answered 1 hour ago
Andy PutmanAndy Putman
31.6k7134214
answered 1 hour ago
Andy PutmanAndy Putman
31.6k7134214
answered 1 hour ago
Andy PutmanAndy Putman
31.6k7134214
answered 1 hour ago
Andy PutmanAndy Putman
31.6k7134214
31.6k7134214
add a comment |
add a comment |
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$begingroup$
Yes, these are matrices with finite order. It's an exercise, not really research level.
$endgroup$
– YCor
1 hour ago